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Register Machines

Register Machines. Connecting evaluators to low level machine code. Plan. Design a central processing unit (CPU) from: wires logic (networks of AND gates, OR gates, etc) registers control sequencer Our CPU will interpret Scheme as its machine language

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Register Machines

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  1. Register Machines • Connecting evaluators to low level machine code

  2. Plan • Design a central processing unit (CPU) from: • wires • logic (networks of AND gates, OR gates, etc) • registers • control sequencer • Our CPU will interpret Scheme as its machine language • Today: Iterative algorithms in hardware • Recursive algorithms in hardware • Then: Scheme in hardware (EC-EVAL) • EC-EVAL exposes more details of scheme than M-EVAL

  3. GCD 24 4 4 Circuit diagram Procedure description 48 30 6 48 30 48 30 Ultimate machine Universal machine 6 6 The ultimate goal (define (gcd a b) ….)

  4. A universal machine • Existence of a universal machine has major implications for what “computation” means • Insight due to Alan Turing (1912-1954) • “On computable numbers with an application to the Entscheidungsproblem, A.M. Turing, Proc. London Math. Society, 2:42, 1937 • Hilbert’s Entscheidungsproblem (decision problem) 1900: Is mathematics decidable? That is, is there a definite method guaranteed to produce a correct decision about all assertions in mathematics? • Church-Turing thesis: Any procedure that could reasonably be considered to be an effective procedure can be carried out by a universal machine (and thus by any universal machine)

  5. Euclid's algorithm to compute GCD (define (gcd a b) (if (= b 0) a (gcd b (remainder a b)))) • Given some numbers a and b • If b is 0, done (the answer is a) • If b is not 0: • the new value of a is the old value of b • the new value of b is the remainder of a  b • start again

  6. test register operation constant button wire Example register machine: datapaths a b = 0 rem t

  7. label operations Example register machine: instructions (controller test-b (test (op =) (reg b) (const 0)) (branch (label gcd-done)) (assign t (op rem) (reg a) (reg b)) (assign a (reg b)) (assign b (reg t)) (goto (label test-b)) gcd-done)

  8. a b = 0 rem t Complete register machine condition sequencer program counter instructions

  9. Datapath components • Button • when pressed, value on input wire flows to output • Register • output the stored value continuously • change value when button on input wire is pressed • Operation • output wire value = some function of input wire values • Test • an operation • output is one bit (true or false) • output wire goes to condition register

  10. 0 1 1 2 3 2 Incrementing a register X • What sequence of button presses will result in the register sum containing the value 2? Y sum 0 1 ? an op thatadds its inputs + ? press sum ? X 0 Y 1 Y 2 X Y Y

  11. Euclid's algorithm to compute GCD (define (gcd a b) (if (= b 0) a (gcd b (remainder a b)))) • Given some numbers a and b • If b is 0, done (the answer is a) • If b is not 0: • the new value of a is the old value of b • the new value of b is the remainder of a  b • start again

  12. X a b Y rem Z t Datapath for GCD (partial) • What sequence of button presses will result in: the register a containing GCD(a,b) the register b containing 0 • The operation rem computes the remainder of a  b press a b t 9 6 ? Z 9 6 3 X 6 6 3 Y 6 3 3 Z 6 3 0 X 3 3 0 Y 3 0 0

  13. Example register machine: instructions (controller test-b (test (op =) (reg b) (const 0)) (branch (label gcd-done)) (assign t (op rem) (reg a) (reg b)) (assign a (reg b)) (assign b (reg t)) (goto (label test-b)) gcd-done)

  14. Instructions • Controller: generates a sequence of button presses • sequencer • instructions • Sequencer: activates instructions sequentially • program counter remembers which one is next • Each instruction: • commands a button press, OR • changes the program counter • called a branch instruction

  15. X Y sum 0 1 + Button-press instructions: the sum example (controller (assign sum (const 0)) <X> (assign sum (op +) (reg sum) (const 1)) <Y> (assign sum (op +) (reg sum) (const 1)))

  16. sequencer: nextPC <- PC + 1 activate instruction at PC PC <- nextPC start again X Y sum 0 1 PC nextPC press 0 1 X 1 2 Y 2 3 1 -- 1 2 Y 2 3 1 -- + Unconditional branch (controller 0 (assign sum (const 0)) increment 1 (assign sum (op +) (reg sum) (const 1)) 2(goto (label increment)))

  17. condition a b = sequencer program counter 0 rem insts t (test (op =) (reg b) (const 0)) (branch (label gcd-done)) gcd-done Conditional branch (controller test-b (assign t (op rem) (reg a) (reg b)) (assign a (reg b)) (assign b (reg t)) (goto (label test-b)) )

  18. Conditional branch details (test (op =) (reg b) (const 0)) • push the button which loads the condition registerfrom this operation's output (branch (label gcd-done)) • Overwrite nextPC register with value if condition register is TRUE • No effect if condition register is FALSE

  19. Datapaths are redundant • We can always draw the data path required for aninstruction sequence • Therefore, we can leave out the data path when describing a register machine

  20. Abstract operations • Every operation shown so far is abstract: • abstract = consists of multiple lower-level operations • Lower-level operations might be: • AND gates, OR gates, etc (hardware building-blocks) • sequences of register machine instructions • Example: GCD machine uses (assign t (op rem) (reg a) (reg b)) • Rewrite this using lower-level operations

  21. Less-abstract GCD machine (controller test-b (test (op =) (reg b) (const 0)) (branch (label gcd-done)) ; (assign t (op rem) (reg a) (reg b)) (assign t (reg a)) rem-loop (test (op <) (reg t) (reg b)) (branch (label rem-done)) (assign t (op -) (reg t) (reg b)) (goto (label rem-loop)) rem-done (assign a (reg b)) (assign b (reg t)) (goto (label test-b)) gcd-done)

  22. Importance of register machine abstraction • A CPU is a very complicated device • We will study only the core of the CPU • eval, apply, etc. • We will use abstract register-machine operations for all the other instruction sequences and circuits: (test (op self-evaluating?) (reg exp)) • remember,(op +) is abstract, (op <) is abstract, etc. • no magic in (op self-evaluating?)

  23. Review of register machines • Registers hold data values • Controller specifies sequence of instructions, order of execution controlled by program counter • Assign puts value into register • Constants • Contents of register • Result of primitive operation • Goto changes value of program counter, and jumps to label • Test examines value of a condition, setting a flag • Branch resets program counter to new value, if flag is true • Data paths are redundant

  24. Machines for recursive algorithms • GCD, odd?, increment • iterative, constant space • factorial, EC-EVAL • recursive, non-constant space • Extend register machines with subroutines and stack • Main points • Every subroutine has a contract • Stacks are THE implementation mechanism for recursive algorithms

  25. Part 1: Subroutines • Subroutine: a sequence of instructions that • starts with a label and ends with an indirect branch • can be called from multiple places • New register machine instructions • (assign continue (label after-call-1)) • store the instruction number corresponding to label after-call-1 in register continue • this instruction number is called the return point • (goto (reg continue)) • an indirect branch • change the PC to the value stored in register continue

  26. (controller (assign (reg sum) (const 0)) (assign continue (label after-call-1)) (goto (label increment)) after-call-1 (assign continue (label after-call-2)) (goto (label increment)) after-call-2 (goto (label done)) increment (assign sum (op +) (reg sum) (const 1)) (goto (reg continue)) done) Example subroutine: increment • set sum to 0, then increment, then increment again • dotted line: subroutineblue: callgreen: labelred: indirect jump

  27. Subroutines have contracts • Follow the contract or register machine will fail: • registers containing input values and return point • registers in which output is produced • registers that will be overwritten • in addition to the output registers increment (assign sum (op +) (reg sum) (const 1)) (goto (reg continue)) • subroutine increment • input: sum, continue • output: sum • writes: none

  28. End of part 1 • Why subroutines? • reuse instructions • reuse data path components • make instruction sequence more readable • just like using helper functions in scheme • support recursion • Contracts • specify inputs, outputs, and registers used by subroutine

  29. 0 a 5 stack b Part 2: Stacks • Stack: a memory device • save a register: send its value to the stack • restore a register: get a value from the stack • When this machine halts, b contains 0: • (controller • (assign a (const 0)) • (assign b (const 5)) • (save a) • (restore b) • )

  30. Stacks: hold many values, last-in first-out • This machine halts with 5 in a and 0 in b (controller 0 (assign a (const 0)) 1 (assign b (const 5)) 2 (save a) 3 (save b) 4 (restore a) 5 (restore b)) contents of stack after step 2 3 4 5 empty 0 5 0 0 • 5 is the top of stack after step 3 • save: put a new value on top of the stack • restore: remove the value at top of stack

  31. 8 5 3 (restore c) (restore b) (restore a) Check your understanding • Draw the stack after step 5. What is the top of stack value? • Add restores so final state is a: 3, b: 5, c: 8, and stack is empty (controller 0 (assign a (const 8)) 1 (assign b (const 3)) 2 (assign c (const 5)) 3 (save b) 4 (save c) 5 (save a) )

  32. Things to know about stacks • stack depth • stacks and subroutine contracts • tail-call optimization

  33. Stack depth • depth of the stack = number of values it contains • At any point while the machine is executing • stack depth = (total # of saves) - (total # of restores) • stack depth limits: • low: 0 (machine fails if restore when stack empty) • high: amount of memory available • max stack depth: • measures the space required by an algorithm

  34. Stacks and subroutine contracts • Standard contract: subroutine increment • input: sum, continue • output: sum • writes: none • stack: unchanged • Rare contract: strange (assign val (op *) (reg val) (const 2)) (restore continue) (goto (reg continue)) • input: val, return point on top of stack • output: val • writes: continue • stack: top element removed

  35. Optimizing tail callsno work after call except (goto (reg continue)) This optimization is important in EC-EVAL • Iterative algorithms expressed as recursive procedures would use non-constant space without it setup Unoptimized version (assign sum (const 15)) (save continue) (assign continue (label after-call)) (goto (label increment)) after-call (restore continue) (goto (reg continue)) setupOptimized version (assign sum (const 15)) (goto (label increment))

  36. End of part 2 • stack • a LIFO memory device • save: put data on top of the stack • restore: remove data from top of the stack • things to know • concept of stack depth • expectations and effect on stack is part of the contract • tail call optimization

  37. Part 3: recursion (define (fact n) (if (= n 1) 1 (* n (fact (- n 1))))) (fact 3) (* 3 (fact 2)) (* 3 (* 2 (fact 1))) (* 3 (* 2 1)) (* 3 2) 6 • The stack is the key mechanism for recursion • remembers return point of each recursive call • remembers intermediate values (eg., n)

  38. (controller (assign continue (label halt)) fact (test (op =) (reg n) (const 1)) (branch (label b-case)) (save continue) (save n) (assign n (op -) (reg n) (const 1)) (assign continue (label r-done)) (goto (label fact)) r-done (restore n) (restore continue) (assign val (op *) (reg n) (reg val)) (goto (reg continue)) b-case (assign val (const 1)) (goto (reg continue)) halt)

  39. Code: base case (define (fact n) (if (= n 1) 1 ...)) fact (test (op =) (reg n) (const 1)) (branch (label b-case)) ... b-case (assign val (const 1)) (goto (reg continue)) • fact expects its input in which register? • fact expects its return point in which register? • fact produces its output in which register? n continue val

  40. Code: recursive call (define (fact n) ... (fact (- n 1)) ...) ... (assign n (op -) (reg n) (const 1)) (assign continue (label r-done)) (goto (label fact)) r-done ... • At r-done, which register will contain the return value of the recursive call? val

  41. Code: after recursive call (define (fact n) ... (* n <return-value> ) ...) (assign val (op *) (reg n) (reg val)) (goto (reg continue)) • Problem! • Overwrote register n as part of recursive call • Also overwrote continue

  42. Code: complete recursive case (save continue) (save n) (assign n (op -) (reg n) (const 1)) (assign continue (label r-done)) (goto (label fact)) r-done (restore n) (restore continue) (assign val (op *) (reg n) (reg val)) (goto (reg continue)) • Save a register if: • value is used after call AND • register is not output of subroutine AND • (register written as part of call OR register written by subroutine)

  43. Check your understanding • Write down the contract for subroutine fact • input: • output: • writes: • stack: n, continue val none unchanged • Writes none? • writes n and continue • but saves them before writing, restores after

  44. Execution trace • Contents of registers and stack at each label • Top of stack at left label continue n val stack • fact halt 3 ??? empty • fact r-done 2 ??? 3 halt • fact r-done 1 ??? 2 r-done 3 halt • b-case r-done 1 ??? 2 r-done 3 halt • r-done r-done 1 1 2 r-done 3 halt • r-done r-done 2 2 3 halt • halt halt 3 6 empty • Contents of stack represents pending operations(* 3 (* 2 (fact 1))) at base case

  45. End of part 3 • To implement recursion, use a stack • stack records pending work and return points • max stack depth = space required • (for most algorithms)

  46. Where we are headed • Next time will use register machine idea to implement an evaluator • This will allow us to capture high level abstractions of Scheme while connecting to low level machine architecture

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